我试图将数据保存到Yii中具有两种不同模型的两个db表。我已经咨询了wiki: http://www.yiiframework.com/wiki/19/how-to-use-a-single-form-to-collect-data-for-two-or-more-models/和http://www.yiiframework.com/forum/index.php/topic/52109-save-data-with-two-models/,但我仍然不能得到数据保存到两个表。我有两个表sales_rep_min_margin
和sales_rep_min_margin_history
:
CREATE TABLE `sales_rep_min_margin` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(32) NOT NULL,
`domestic` int(2) NOT NULL,
`overseas` int(2) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
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历史表:
CREATE TABLE `sales_rep_min_margin_history` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`min_margin_id` int(11) NOT NULL,
`from` int(11) DEFAULT NULL,
`to` int(11) DEFAULT NULL,
`update_username` varchar(32) NOT NULL,
`update_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`id`),
KEY `min_margin_id` (`min_margin_id`),
CONSTRAINT `sales_rep_min_margin_history_ibfk_1` FOREIGN KEY (`min_margin_id`) REFERENCES `sales_rep_min_margin` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
My SalesRepMinMarginController code (right now) is:
public function actionCreate() {
$model = new SalesRepMinMargin;
$model2 = new SalesRepMinMarginHistory;
//Uncomment the following line if AJAX validation is needed
//$this->performAjaxValidation($model);
if (isset($_POST['SalesRepMinMargin'])) {
$model->attributes = $_POST['SalesRepMinMargin'];
if ($model->save())
$this->redirect(array('view', 'id' => $model->id));
}
if (isset($_POST['SalesRepMinMarginHistory'])) {
$model2->attributes = $_POST['SalesRepMinMarginHistory'];
$model2->save();
$this->render('create', array(
'model' => $model,
));
}
}
and 'SalesRepMinMarginHistoryController':
public function actionUpdate($id)
{
$model=$this->loadModel($id);
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['SalesRepMinMarginHistory']))
{
$model->attributes=$_POST['SalesRepMinMarginHistory'];
if($model->save())
$this->redirect(array('view','id'=>$model->id));
}
$this->render('update',array(
'model'=>$model,
));
}
我只需要将数据保存到表中,但我不需要视图中'history'表的数据。任何帮助是非常感谢!有人给了我下面的代码,但是,它不起作用:
public function actionCreate() {
$model = new SalesRepMinMargin;
$model2 = new SalesRepMinMarginHistory;
//Uncomment the following line if AJAX validation is needed
//$this->performAjaxValidation($model);
if (isset($_POST['SalesRepMinMargin'])) {
$model->attributes = $_POST['SalesRepMinMargin'];
if ($model->save()) {
if (isset($_POST['SalesRepMinMarginHistory'])) {
$model2->attributes = $_POST['SalesRepMinMarginHistory'];
$model2->save();
}
$this->redirect(array('view', 'id' => $model->id));
}
}
$this->render('create', array(
'model' => $model, 'model2' => $model2,
));
}
首先使用模型
在第一个表中保存数据$model = new SalesRepMinMargin;
$model->attributes = $_POST['SalesRepMinMargin'];
if($model->save(false)) {
/* use second model */
$model2 = new SalesRepMinMarginHistory;
$model2->attributes = $_POST['SalesRepMinMarginHistory'];
$model2->save();
$this->redirect(array('view', 'id' => $model->id));
}
第二个模型的保存函数不受影响,因为在保存第一个模型后会调用重定向函数。
查看wiki页面。
if($valid)
{
// use false parameter to disable validation
$a->save(false);
$b->save(false);
// ...redirect to another page
}
应该在成功保存两个模型后调用重定向函数
我不是专家,但也许您可以创建一个触发器来实现更改?当您在sales_rep_min_margin表中获得一个新表时,您也可以将数据转储到sales_rep_min_margin_history表中。